This is a tedious drawing so if you are going to procede I'll thank you in advance. Here is my reasoning for solving this problem.Draw an isosceles triangle ABC with Side AB = Side AC. Draw a line from C to side AB and label that line CD. Now draw a line from B to side AC. Label that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE.

Find what angle EDC is by using geometry only and no trigonometry.

Don't cheat and use protractors/rulers/all that stuff either! Go by pure geometric reasoning

Call the intersection of BE and CD point M. Let ADE=x, EDM=y, DEA=z, and DEM=140-z.

Thus we have the following system of equations:

(x+y)+20+10=180

x+20+z=180

50+y+(140-z)=180

And when I solve this I get a large angle, a negative angle, and an angle of zero. I think my work is sound. Please look it over.

Thanks